Problem D: ShellSort
He made each turtle stand on another one‘s back
And he piled them all up in a nine-turtle stack.
And then Yertle climbed up. He sat down on the pile.
What a wonderful view! He could see ‘most a mile!
The Problem
King Yertle wishes to rearrange his turtle throne to place his highest-ranking nobles and closest advisors nearer to the top. A single operation is available to change the order of the turtles in the stack: a turtle
can crawl out of its position in the stack and climb up over the other turtles to sit on the top.
Given an original ordering of a turtle stack and a required ordering for the same turtle stack, your job is to determine a minimal sequence of operations that rearranges the original stack into the required stack.
The first line of the input consists of a single integer K giving the number of test cases. Each test case consist on an integer n giving the number of turtles in the stack. The next n lines
specify the original ordering of the turtle stack. Each of the lines contains the name of a turtle, starting with the turtle on the top of the stack and working down to the turtle at the bottom of the stack. Turtles have unique names, each of which is a string
of no more than eighty characters drawn from a character set consisting of the alphanumeric characters, the space character and the period (`.‘). The next n lines in the input gives the desired ordering of the stack, once again by naming turtles from
top to bottom. Each test case consists of exactly 2n+1 lines in total. The number of turtles (n) will be less than or equal to two hundred.
For each test case, the output consists of a sequence of turtle names, one per line, indicating the order in which turtles are to leave their positions in the stack and crawl to the top. This sequence of operations
should transform the original stack into the required stack and should be as short as possible. If more than one solution of shortest length is possible, any of the solutions may be reported. Print a blank line after each test case.
Sample Input
2 3 Yertle Duke of Earl Sir Lancelot Duke of Earl Yertle Sir Lancelot 9 Yertle Duke of Earl Sir Lancelot Elizabeth Windsor Michael Eisner Richard M. Nixon Mr. Rogers Ford Perfect Mack Yertle Richard M. Nixon Sir Lancelot Duke of Earl Elizabeth Windsor Michael Eisner Mr. Rogers Ford Perfect Mack
Sample Output
Duke of Earl Sir Lancelot Richard M. Nixon Yertle
这是一道模拟题，基本思想是对于在目标顺序里的每一只乌龟从下往上查找在原顺序中的同一乌龟所在位置，查找过程中原顺序的未匹配的乌龟均需要记录下来（如果该乌龟能查找到），每一次查找都需要在原顺序的匹配的乌龟的下一个开始，如此进行直到查找不到，记下目标顺序中第一个查找不到的乌龟，在目标顺序中从此乌龟开始，查找记录中是否有，若有则将其放到原顺序的最上面，并进行目标顺序中的下一个查找，若没有了继续下一波查找，直到最终顺序满足要求为止。
#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
using namespace std;
vector<string> v1;
vector<string> v2;
vector<string> v;
string s1, s2;
int size;
int n;
int OneTraverse(void);
int main(void){
string s1, s2;
int size;
int n;
#ifndef ONLINE_JUDGE
freopen("f://infile.txt", "r", stdin);
#endif
cin >> size;
for(int i = 0; i < size; i++){
cin >> n;
cin.ignore(100, '\n');
v1.clear();
v2.clear();
for(int j = 0; j < n; j++){
// cin.ignore(100, '\n');
getline(cin, s1);
v1.insert(v1.begin(), s1);
}
for(int j = 0; j < n; j++){
// cin.ignore(100, '\n');
getline(cin, s2);
v2.insert(v2.begin(), s2);
}
int s = 0;
while(1){
if(OneTraverse()){
break;
}
}
cout << endl;
}
return 0;
}
int OneTraverse(void){
v.clear();
size_t k;
size_t m;
size_t preM = 0;
vector<string> tempV;
for(k = 0; k < v2.size(); k++){
tempV.clear();
for(m = preM; m < v1.size(); m++){
if(v1[m] == v2[k]){
preM = m+1;
break;
}
else{
tempV.push_back(v1[m]);
}
}
if(m < v1.size())
v.insert(v.end(), tempV.begin(), tempV.end());
else
break;
}
if(v.size() == 0)
return 1;
size_t kk;
for(kk = k; kk < v2.size(); kk++){
size_t index;
for(index = 0; index < v.size(); index++){
if(v[index] == v2[kk]){
string temp;
temp = v[index];
v1.erase(find(v1.begin(), v1.end(), temp));
v1.push_back(temp);
cout << temp << endl;
break;
}
}
if(index == v.size())
break;
}
return 0;
}
uva10152-ShellSort